Question

1、已知sinθ+cosθ=2sinα,sinθcosθ=sin∧2(β),求证：4cos∧2(2α)=cos∧2(2β

Answer 1

1.sinθ+cosθ=2sinα...①

sinθcosθ=sin^2(β)...②

①^2-2②=4sin^2(α)-2sin^2(β)=1

则2(1-cos2α)+(cos2β-1)=1

整理得cos2β=2cos2α,两边平方即得4cos^2(2α)=cos^2(2β)

2.f(x)=(√3)sin(2x)+2cos^2(x)+m=(√3)sin(2x)+cos(2x)+m+1

=2(sin(2x+π/6)+m+1,在区间[0,π/2]上最大值为2+m+1=6,m=3

最小值为m+1-1/2=7/2